If \(\mathrm{f}^{\prime}(x)=\tan ^{-1}(\sec x+\tan x),-\frac{\pi}{2} < x < \frac{\pi}{2}\) and \(f(0)=0\), then \(f(1)\) is

\(f^{\prime}(x) =\tan ^{-1}(\sec x+\tan x) \)
\( =\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right) \)
\( =\tan ^{-1}\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}\right] \)
\( =\tan ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right) \)
\( =\tan ^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right) \)
\( =\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]\)
\(\therefore \quad \mathrm{f}^{\prime}(x) =\frac{\pi}{4}+\frac{x}{2} \)
\( \Rightarrow \mathrm{f}(x) =\int\left(\frac{\pi}{4}+\frac{x}{2}\right) \mathrm{d} x \)
\( =\frac{\pi x}{4}+\frac{1}{2} \cdot \frac{x^2}{2}+\mathrm{c}\) \(\therefore \quad f(0)=c \)
\(\Rightarrow c=0 \ldots[\because f(0)=0 \text { (given) }]\)
\(\therefore \mathrm{f}(x)=\frac{\pi x}{4}+\frac{x^2}{4} \)
\( \Rightarrow \mathrm{f}(1)=\frac{\pi+1}{4}\)